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ch21

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CHAPTER 21 Numerics for ODEs and PDEs Major Changes These include automatic variable step size selection in modern codes, the discussion of the Runge–Kutta–Fehlberg method, backward Euler’s method and its application to stiff ODEs, and the extension of Euler and Runge–Kutta methods to systems and higher order equations. SECTION 21.1. Methods for First-Order ODEs, page 886 Purpose. To explain three numerical methods for solving initial value problems y 9 5 ƒ( x , y ), y ( x 0 ) 5 y 0 by stepwise computing approximations to the solution at x 1 5 x 0 1 h , x 2 5 x 0 1 2 h , etc. Main Content, Important Concepts Euler’s method (3) Automatic variable step size selection Improved Euler method (8), Table 21.2 Classical Runge–Kutta method (Table 21.4) Error and step size control Runge–Kutta–Fehlberg method Backward Euler’s method Stiff ODEs Comments on Content Euler’s method is good for explaining the principle but is too crude to be of practical value. The improved Euler method is a simple case of a predictor–corrector method. The classical Runge–Kutta method is of order h 4 and is of great practical importance. Principles for a good choice of h are important in any method. ƒ in the equation must be such that the problem has a unique solution (see Sec. 1.7). SOLUTIONS TO PROBLEM SET 21.1, page 897 2. y 5 e x . Computed values are x n y n y ( x n ) Error Error in Prob. 1 0.01 1.010000 1.010050 0.000050 0.02 1.020100 1.020201 0.000101 0.03 1.030301 1.030455 0.000154 0.04 1.040604 1.040811 0.000207 0.05 1.051010 1.051271 0.000261 0.06 1.061520 1.061837 0.000317 0.07 1.072135 1.072508 0.000373 0.08 1.082857 1.083287 0.000430 0.09 1.093685 1.094174 0.000489 0.10 1.104622 1.105171 0.000549 0.005171 329 im21.qxd 9/21/05 1:41 PM Page 329

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We see that the error of the last value has decreased by a factor 10, owing to the smaller step. In most cases the method will be too inaccurate. 4. This is a special Riccati equation. Set y 1 x 5 u , then u 9 5 u 2 1 1 and u 9 /( u 2 1 1) 5 1. By integration, arctan u 5 x 1 c and u 5 tan ( x 1 c ) 5 y 1 x so that y 5 tan ( x 1 c ) 2 x and c 5 0 from the initial condition. Hence y 5 tan x 2 x . The calculation is x n y n y ( x n ) Error 0.1 0.000000 0.000335 0.000335 0.2 0.001000 0.002710 0.001710 0.3 0.005040 0.009336 0.004296 0.4 0.014345 0.022793 0.008448 0.5 0.031513 0.046302 0.014789 0.6 0.059764 0.084137 0.024373 0.7 0.103292 0.142288 0.038996 0.8 0.167820 0.289639 0.061818 0.9 0.261488 0.360158 0.098670 1.0 0.396393 0.557408 0.161014 Although the ODE is similar to that in Prob. 3, the error is greater by about a factor 10. This is understandable because tan x becomes infinite as x * 1 _ 2 p . 6. The solution is y 5 1/(1 1 4 e 2 x ). The 10S-computation, rounded to 6D, gives x n y n y ( x n ) Error y ( x n ) 2 y n 0.1 0.216467 0.216481 0.000014 0.2 0.233895 0.233922 0.000027 0.3 0.252274 0.252317 0.000043 0.4 0.271587 0.271645 0.000058 0.5 0.291802 0.291875 0.000073 0.6 0.312876 0.312965 0.000089 0.7 0.334754 0.334858 0.000104 0.8 0.357366 0.357485 0.000119 0.9 0.380633 0.380767 0.000134 1.0 0.404462 0.404609 0.000147 8. The solution is y 5 3 cos x 2 2 cos 2 x . The 10S-computation rounded to 6D gives x n y n y ( x n ) Error y ( x n ) 2 y n 0.1 1.00492 1.00494 0.00002 0.2 1.01900 1.01914 0.00014 0.3 1.04033 1.04068 0.00035 0.4 1.06583 1.06647 0.00064 0.5 1.09140 1.09245 0.00105 (continued) 330 Instructor’s Manual im21.qxd 9/21/05 1:41 PM Page 330
x n y n y ( x n ) Error y ( x n ) 2 y n 0.6 1.11209 1.11365 0.00156 0.7 1.12237 1.12456 0.00219 0.8 1.11636 1.11932 0.00296 0.9 1.08817 1.09203 0.00386 1.0 1.03212 1.03706 0.00494 10. The solution is y 5 2 tanh 1 _ 2 x . The 10S-computation rounded to 6D gives x n y n y ( x n ) Error y ( x n ) 2 y n 0.1 0.099875 0.099917 0.000042 0.2 0.199252 0.199336 0.000084 0.3 0.297644 0.297770 0.000126 0.4 0.394582 0.394750 0.000168 0.5 0.489626 0.489838 0.000212 0.6 0.582372 0.582626 0.000254 0.7 0.672455 0.672752 0.000297 0.8 0.759561 0.759898 0.000337 0.9 0.843421 0.843798 0.000377 1.0 0.923819 0.924234 0.000415 The error is growing relatively slowly, because the solution itself remains less than 2.

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ch21 - im21.qxd 1:41 PM Page 329 CHAPTER 21 Numerics for...

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